Prove that $R$ is an equivalence relation. Define the following $R$ on the set of integers $\Bbb{Z}$
$(a,b) \in R$ if and only if $3a + 5b$ is divisible by $8$
Prove that $R$ is an equivalence relation.
Attempt:
Reflexive:
a~a if and only if $3a+5a$ is divisible by 8
Since $3a+5a = 8a$ is divisible by 8, the relationship is reflexive.
Symmetry:
Let $a,b,k \in \Bbb{Z} $
$aRb$ = $ 3a+5b = 8k$
$bRa$ = $ 5b + 3a = 8k$
Therefore, it's symmetric
Transitivity:
Let $a,b,c, i, k \in \Bbb{Z} $
$aRb$ = $ 3a+5b = 8i$
$bRc$ = $ 5b + 3c = 8k$
Now what?
 A: The proof of symmetry is not correct. You are given that $3a+5b$ is a multiple of $8$. You need to show that $3b+5a$ is a multiple of $8$.
Note that $(3a+5b)+(3b+5a)=8(a+b)$. So if $3a+5b=8k$, then $3b+5a=8(a+b)-8k$. The right side is clearly divisible by $8$.
For transitivity, it's your turn. You need to show that if $3a+5b$ is divisible by $8$, and $3b+5c$ is divisible by $8$, then $3a+5c$ is divisible by $8$. Hint: add.
Another way: The following is a slightly more conceptual way of looking at things. Note that $3a+5b=3a-3b+8b$. So $3a+5b$ is divisible by $8$ if and only if $3a-3b$ is divisible by $8$. Now verifying we have an equivalence relation is much easier. For example, for symmetry, it is clear that if $3a-3b$ is divisible by $8$, then $3b-3a$ is divisible by $8$.
A: If $f:X\rightarrow Y$ is a function then the relation $\sim$ defined
by $a\sim b\iff f\left(a\right)=f\left(b\right)$ is automatically
an equivalence relation on $X$. This is easy to verify.
In your case we have:
$aRb\iff8\mid3a+5b\iff8\mid3a-3b\iff8\mid3\left(a-b\right)\iff8\mid a-b\iff\nu\left(a\right)=\nu\left(b\right)$
where $\nu$ is the (natural) function $\nu:\mathbb{Z}\rightarrow\mathbb{Z}/8\mathbb{Z}$
defined by $a\mapsto a+8\mathbb{Z}$.
Proved is now that $R$ is an equivalence relation.
A: Now, well, you want to show transitivity. Your beginning is not correct, So suppose $aRb$ and $bRC$. We want to show $aRc$. We have
$ 3a + 5b = 8i $ and $3b + 5c= 8j$. Notice $ 3b = 8j - 5c = 8i - 3a - 2b $
$$\therefore 3a - 5c = 8i  - 8j - 2b \implies 3a + 5c = 8i - 8j -2b + 10c$$
$$ \implies 3a + 5c = 8i - 8j -2b + (8j - 6b) =8i  - 8b = 8(i - b)$$
Hence, $aRc$
A: This problem is easier if you note:
$$aRb \equiv 3a + 5b = 0 \equiv 3a = -5b \equiv 3a = 3b \equiv a = b \pmod 8$$
Transitive:
$$  aRa $$
$$  a = a \pmod 8 $$
Reflexive
$$  aRb \equiv bRa$$
$$  a = b \equiv b = a \pmod 8$$
Transitive
$$  \left( aRb \land bRc \right) \equiv aRc $$
$$  \left( a = b \land b = c \right) \equiv a = c \pmod 8 $$
